2000 AMC 10 Problems/Problem 16

Problem

The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$ . Find the length of segment $AE$ .

$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); [/asy]$

$\mathrm{(A)}\ \frac{4\sqrt{5}}{3} \qquad\mathrm{(B)}\ \frac{5\sqrt{5}}{3} \qquad\mathrm{(C)}\ \frac{12\sqrt{5}}{7} \qquad\mathrm{(D)}\ 2\sqrt{5} \qquad\mathrm{(E)}\ \frac{5\sqrt{65}}{9}$

Solution

Solution 1

Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$ . If we set the bottom left point at $(0,0)$ , then $A=(0,3)$ , $B=(6,0)$ , $C=(4,2)$ , and $D=(2,0)$ .

The line $l_1$ is given by the equation $y=m_1x+b_1$ . The $y$ -intercept is $A=(0,3)$ , so $b_1=3$ . We are given two points on $l_1$ , hence we can compute the slope, $m_1$ to be $\frac{0-3}{6-0}=\frac{-1}{2}$ , so $l_1$ is the line $y=\frac{-1}{2}x+3$

Similarly, $l_2$ is given by $y=m_2x+b_2$ . The slope in this case is $\frac{2-0}{4-2}=1$ , so $y=x+b_2$ . Plugging in the point $(2,0)$ gives us $b_2=-2$ , so $l_2$ is the line $y=x-2$ .

At $E$ , the intersection point, both of the equations must be true, so $\begin{align*} y=x-2, y=\frac{-1}{2}x+3 &\Rightarrow x-2=\frac{-1}{2}x+3 \\ &\Rightarrow x=\frac{10}{3} \\ &\Rightarrow y=\frac{4}{3} \\ \end{align*}$

We have the coordinates of $A$ and $E$ , so we can use the distance formula here: $\sqrt{\left(\frac{10}{3}-0\right)^2+\left(\frac{4}{3}-3\right)^2}=\frac{5\sqrt{5}}{3}$

which is answer choice $\boxed{\text{B}}$

Solution 2

$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2.5,.5),E); draw((6,0)--(4,2)); draw((0,3)--(2.5,.5)); [/asy]$

Draw the perpendiculars from $A$ and $B$ to $CD$ , respectively. As it turns out, $BC \perp CD$ . Let $F$ be the point on $CD$ for which $AF\perp CD$ .

$m\angle AFE=m\angle BCE=90^\circ$ , and $m\angle AEF=m\angle CEB$ , so by AA similarity, $\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}$

By the Pythagorean Theorem, we have $AB=\sqrt{3^2+6^2}=3\sqrt{5}$ , $AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}$ , and $BC=\sqrt{2^2+2^2}=2\sqrt{2}$ . Let $AE=x$ , so $BE=3\sqrt{5}-x$ , then $\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}$ $x=\frac{5\sqrt{5}}{3}$

This is answer choice $\boxed{\text{B}}$

Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.

Solution 3

$[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$

Drawing line $\overline{BD}$ and parallel line $\overline{CF}$ , we see that $\triangle FCE \sim \triangle BDE$ by AA similarity. Thus $\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}$ . Reciprocating, we know that $\frac{EB}{FE} = 2$ so $\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3$ . Reciprocating again, we have $\frac{FE}{FB} = \frac{1}{3} \Rightarrow FE = \frac{1}{3}FB$ . We know that $FD = 2$ , so by the pythagorean theorem, $FB = \sqrt{2^{2} + 4^{2}} = 2\sqrt{5}$ . Thus $FE = \frac{1}{3}FB = \frac{2\sqrt{5}}{3}$ . Applying the pythagorean theorem again, we have $AF = \sqrt{1^{2}+2^{2}} = \sqrt{5}$ . We finally have $AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

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